Verifying this proof: $a_n$ is cauchy $\implies a^2_n$ is cauchy Prove: $a_n$ is cauchy $\implies a^2_n$ is cauchy
Proof:
Let $a_n$ be a cauchy sequence. Then we know:
$\forall_{\epsilon > 0} \exists N_1 s.t. \forall_{n, m} \implies |a_n - a_m| < \epsilon$  We will set $\epsilon$ to some value later to prove the statement
Consider $a^2_n$:
$\forall_{\epsilon >0}$ take $N = N_1$ then $\forall_{n,m} > N:$
$|a^2_n - a^2_m| = |(a_n - a_m)(a_n + a_m)|$
From the above, take $\epsilon = \dfrac{\epsilon}{2(a_m + a_n)}$ Then we have:
$|\dfrac{\epsilon}{2(a_m+a_n)}(a_m+a_n)| = \dfrac{\epsilon}{2} < \epsilon$
End of Proof
 A: There are several problems here. First, the statement $$\forall_{\epsilon > 0} \exists N_1 s.t. \forall_{n, m} \implies |a_n - a_m| < \epsilon$$ is meaningless as it stands: the implication has no premiss. It should have been
$$\forall\epsilon>0\exists N\forall n,m\Big(n,m\ge N\implies|a_n-a_m|<\epsilon\Big)$$
or simply
$$\forall\epsilon>0\exists N\forall n,m\ge N\Big(|a_n-a_m|<\epsilon\Big)\;.$$
Secondly, you don’t get to tinker with $\epsilon$: you’re given an $\epsilon$, and you have to find an $N$ that makes the statement $$\forall n,m\ge N\Big(|a_n^2-a_m^2|<\epsilon\Big)$$ true.
Thirdly, you can’t choose $N$ in a way that depends on $n$ and $m$.
By hypothesis you know that for each $\epsilon>0$ there’s an $N_\epsilon$ such that $|a_n-a_m|<\epsilon$ whenever $n,m\ge N_\epsilon$. You’ve also correctly noticed that $|a_n^2-a_m^2|=|a_n-a_m||a_n+a_m|$. Thus, you know that $$|a_n^2-a_m^2|<\epsilon|a_n+a_m|\tag{1}$$ whenever $n,m\ge N$. Suppose that you knew that there was some bound $M$ such that $|a_n+a_m|<M$ for all $n,m\in\Bbb N$. Then you could infer from $(1)$ that $|a_n^2-a_m^2|<\epsilon M$ for all $n,m\ge N_\epsilon$. What’s more, this calculation works for every $\epsilon>0$. Thus, when given some $\epsilon>0$, you could set $N=N_{\epsilon/M}$ and conclude that $$|a_n^2-a_m^2|<\frac{\epsilon}M|a_n+a_m|<\left(\frac{\epsilon}M\right)M=\epsilon$$ whenever $n,m\ge N$. You would then have the required recipe for finding $N$ given $\epsilon$. It only remains, then, to show that there really is some $M$ such that $|a_n+a_m|<M$ for all $n,m\in\Bbb N$.
Notice that it’s enough to find an $M_1$ such that $|a_n|<M_1$ for all $n\in\Bbb N$, for then $$|a_n+a_m|\le|a_n|+|a_m|<2M_1$$ for all $n,m\in\Bbb N$, and we can simply take $M=2M_1$. That is, it’s enough to show that every Cauchy sequence is bounded. This isn’t too hard; I’ll leave it to you, but feel free to ask for more help if you get completely stuck.
